- #1

Oster

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I am not too confident about my solution. Can someone please look through it?

Fortunate guess:-

The set J(y) = {f in C[0,1] / f(y)=0} where y is fixed in [0,1]

This is an ideal because

(1) If g,h belong to J(y) then g-h belongs to J(y) => (J(y),+) is a subgroup of (C[0,1],+)

(2) If g is in J(y) and h is in C[0,1] then g.h is in J(y)

Also, C[0,1]/J(y) is a field because,

If J(y)+f is in C[0,1]/J(y) and J(y)+f =/= J(y), then f(y) =/= 0.

Let g(x)=1/f(y) for all x in C[0,1]

Then, (J(y)+g).(J(y)+f)=(J(y)+f.g)=(J(y)+1) [because f.g-1 is in J(y)]

Hence (J(y)+f) is invertible.

And, C[0,1]/J(y) is a field => J(y) is a maximal ideal.

To show there are no other maximal ideals:-

Let I be an ideal of C[0,1] such that I is not {0} and I is not a subset of any J(y)

That implies, for each z in [0,1], there exists f in I such that f(z) =/= 0.

multiplying this f by an appropriate constant function AND an appropriate continuous piecewise linear function will guarantee, for each z, the existence of a function g in I such that g(z)>1 AND g>0 on [0,1].

for z in [0,1], the corresponding function g and e=1/2, there exists a number d(z)>0 such that in B(z,d(z)) g is always greater than 1/2.

B(z,d(z)) is an open cover for [0,1] so it must admit a finite subcover. This will give me a finite number of functions g1, g2,..., gk in I such that their sum is a function in I that never becomes zero. Multiplying this function by its reciprocal will imply that the identically 1 function is in I. Which implies I=C[0,1]

Hence the J(y)'s are the only maximal ideals in C[0,1] (?)

Thank you for reading the whole thing.